Aufgaben:Exercise 2.4: Dual Code and Gray Code

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Quaternary signals with dual and Gray coding

The two shown signals  $s_{1}(t)$  and  $s_{2}(t)$  are two different realizations of a redundancy-free quaternary transmitted signal,  both derived from the blue drawn binary source signal  $q(t)$. 

For one of the transmitted signals,  the so-called  dual code  with mapping

$$\mathbf{LL}\hspace{0.1cm}\Leftrightarrow \hspace{0.1cm} -s_0, \hspace{0.35cm} \mathbf{LH}\hspace{0.1cm}\Leftrightarrow \hspace{0.1cm} -s_0/3,\hspace{0.35cm}\mathbf{HL}\hspace{0.1cm}\Leftrightarrow \hspace{0.1cm} +s_0/3, \hspace{0.35cm} \mathbf{HH}\hspace{0.1cm}\Leftrightarrow \hspace{0.1cm} +s_0$$

was used,  for the other one a certain form of a  Gray code.  This is characterized by the fact that the binary representation of adjacent amplitude values always differ only in a single bit.

The solution of the exercise should be based on the following assumptions:

• The amplitude levels are  $±3\, \rm V$  and  $±1 \, \rm V$.

• The decision thresholds lie in the middle between two adjacent amplitude values,  i.e. at  $–2\, \rm V$,  $0\, \rm V$  and  $+2\, \rm V$.

• The noise rms value  $\sigma_{d}$  is to be chosen so that the falsification probability from the outer symbol  $(+s_0)$  to the nearest symbol  $(+s_{0}/3)$  is exactly  $p = 1\%$.

• Falsification to non-adjacent symbols can be excluded;  in the case of Gaussian perturbations,  this simplification is always allowed in practice.

One distinguishes in principle between

• the  "symbol error probability"  $p_{\rm S}$  (related to the quaternary signal)  and
• the  "bit error probability"  $p_{B}$  (related to the binary source signal).

Notes:

• The exercise is part of the chapter  "Basics of Coded Transmission".

• Reference is also made to the chapter  "Redundancy-Free Coding".

• For numerical evaluation of the Q–function you can use the HTML5/JavaScript applet  "Complementary Gaussian Error Functions".

Questions

1 Which of the signals  $s_{1}(t)$  or  $s_{2}(t)$  uses  Gray coding?

	$s_{1}(t)$  uses Gray coding.
	$s_{2}(t)$  uses Gray coding.

2 Determine the noise rms value from the given condition.

$\sigma_{d} \ = \ $ $\ \rm V$

3 What is the symbol error probability using the  Gray code?

$p_{\rm S} \ = \ $ $\ \%$

4 What is the bit error probability with the Gray code?

$p_{\rm B} \ = \ $ $\ \%$

5 What is the symbol error probability with the  dual code?

$p_{\rm S} \ = \ $ $\ \%$

6 What is the bit error probability with the dual code?

$p_{\rm B} \ = \ $ $\ \%$

Solution

Solution

(1)  In the signal  $s_{2}(t)$  one recognizes the realization of the dual code indicated at the beginning.  On the other hand,  in the signal  $s_{2}(t)$  a Gray code   $\Rightarrow$   solution 1 with the following mapping was used:

$$\mathbf{HH}\hspace{0.1cm}\Leftrightarrow \hspace{0.1cm} -1, \hspace{0.35cm} \mathbf{HL}\hspace{0.1cm}\Leftrightarrow \hspace{0.1cm} -1/3, \hspace{0.35cm} \mathbf{LL}\hspace{0.1cm}\Leftrightarrow \hspace{0.1cm} +1/3, \hspace{0.35cm} \mathbf{LH}\hspace{0.1cm}\Leftrightarrow \hspace{0.1cm} +1 \hspace{0.05cm}.$$

(2)  Let the probability  $p$  that the amplitude value  $3 \, \rm V$  falls below the adjacent decision threshold  $2\, \rm V$  due to the Gaussian distributed noise with standard deviation  $\sigma_{d}$  be $1\, \%$.  It follows that:

$$ p = {\rm Q} \left ( \frac{3\,{\rm V} - 2\,{\rm V}} { \sigma_d}\right ) = 1 \%\hspace{0.3cm}\Rightarrow \hspace{0.3cm} {1\,{\rm V} }/ { \sigma_d} \approx 2.33 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} { \sigma_d}\hspace{0.15cm}\underline {\approx 0.43\,{\rm V}}\hspace{0.05cm}.$$

(3)  The two outer symbols are each falsified with probability  $p$,  the two inner symbols with double probability  $(2p)$.  By averaging considering equal symbol occurrence probabilities,  we obtain

$$p_{\rm S} = 1.5 \cdot p \hspace{0.15cm}\underline { = 1.5 \,\%} \hspace{0.05cm}.$$

(4)  Each symbol error results in exactly one bit error.  However,  since each quaternary symbol contains exactly two binary symbols,  the bit error probability is obtained:

$$p_{\rm B} = {p_{\rm S}}/ { 2}\hspace{0.15cm}\underline { = 0.75 \,\%} \hspace{0.05cm}.$$

(5)  When calculating the symbol error probability  $p_{\rm S}$,  the mapping used is not taken into account.  As in subtask  (3),  we obtain  $p_{\rm S} \hspace{0.15cm}\underline{ = 1.5 \, \%}$.

(6)  The two outer symbols are falsified with  $p$  and lead to only one bit error each even with dual code.

• The inner symbols are falsified with  $2p$  and now lead to  $1.5$  bit errors on average.
• Taking into account the factor  $2$  in the denominator – see subtask  (2)  – we thus obtain for the bit error probability of the dual code:

$$p_{\rm B} = \frac{1} { 4} \cdot \frac{p + 2p \cdot 1.5 + 2p \cdot 1.5 + p} { 2} = p \hspace{0.15cm}\underline { = 1 \,\%} \hspace{0.05cm}.$$